The tensor denoted $X\otimes df$ acts as the rank-one <endomorphism> $Y\mapsto df(Y)X=Y(f)X$. In the paper's covector-first ordering it is written $df\otimes X$, using the canonical interchange of tensor factors; this is the same <endomorphism>.

On any <smooth function> $h$,
$$
(f\mathcal L_X-\mathcal L_{fX})h=fX(h)-(fX)(h)=0.
$$
On a <vector field> $Y$, the <Lie bracket of vector fields> gives
$$
[fX,Y]=f[X,Y]-Y(f)X,
$$
so
$$
(f\mathcal L_X-\mathcal L_{fX})Y=Y(f)X=D_{X\otimes df}Y.
$$
Multiplication by $f$ preserves the tensor-product <Leibniz rule> and commutation with <tensor contractions>, since the latter are linear over <smooth functions>. Hence the difference is a real-linear contraction-compatible <tensor derivation> whose scalar operator is zero. Its agreement with $D_{X\otimes df}$ on functions and <vector fields> determines its action on every tensor by uniqueness:
$$
\boxed{D_{X\otimes df}=f\mathcal L_X-\mathcal L_{fX}.}
$$
For example, on a <differential one-form> this gives $-\omega(X)df$. Equivalently $\mathcal L_{fX}\omega=f\mathcal L_X\omega+\omega(X)df$, which checks the sign independently. This is the <scaled Lie derivative defect identity>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 115 / 2 / d / Solution

{title2=$f\mathcal L_X-\mathcal L_{fX}=D_{X\otimes df}$}

For a <smooth function> $f$ and <vector field> $X$,
$$
f\mathcal L_X-\mathcal L_{fX}=D_{X\otimes df}
$$
on every <tensor field>, with $X\otimes df$ interpreted as the <endomorphism> $Y\mapsto Y(f)X$. Both sides vanish on functions; on <vector fields> this follows from $[fX,Y]=f[X,Y]-Y(f)X$. Both are contraction-compatible <tensor derivations>, so agreement on functions and <vector fields> proves equality on all tensor types. In particular $\mathcal L_{fX}\omega=f\mathcal L_X\omega+\omega(X)df$.

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